# Lorenz formulas

1. Jun 7, 2005

### Moe_the_Genius

In the time lorenz formulas, how can u tell what the difference between t-zero and T is?
t-zero
T= --------------------
-------------------
| 1 - (v^2/c^2)
\|

for example in the following problem:
With what speed will a clock have to be moving in order to run at rate that is one-half the rate of a clock at rest?

How can u know what to plug in for t-zero and T?

Thanks so much

2. Jun 7, 2005

### pmb_phy

I have always found it very helpful in these cases for follow the derivation of the formula. Then you'll understand it much better. See - http://www.geocities.com/physics_world/sr/light_clock.htm

Pete

3. Jun 7, 2005

### dodo

Hint: your T and t-zero are not points in time, but time intervals.

4. Jun 7, 2005

### Moe_the_Genius

thank u for the tips yet still I get the wrong number when I plug my numbers in:

-->With what speed will a clock have to be moving in order to run at rate that is one-half the rate of a clock at rest?

For the problem, I plug in T=(1/2)t and t-zero=t, because I think the moving clock the time is one half-t, while to the clock at rest the time is t. Doesn't t-0 represent what the clock at rest observes??

thanks for any further help anyone provides

5. Jun 7, 2005

### Staff: Mentor

$T_0$ represents the time as measured by the "moving" clock; $T$ represents the time as measured by "stationary" clocks.

$$T = \frac{T_0}{\sqrt{1 - v^2/c^2}}$$

In your example, if the moving clock measures t seconds, then the stationary observers will measure 2t seconds according to their clocks. (They measure the moving clock to be running at half the normal rate.)

6. Jun 7, 2005

### robphy

This mysterious and often confusing formula of time-dliation can interpreted geometrically.
As Dodo says, "Hint: your T and t-zero are not points in time, but time intervals."

Let's introduce events:
the common meeting event O,
the distant event t0 on the moving clock [when the moving clock reads t0],
the local event T [when the stationary clock reads T], which our stationary observer says is simultaneous with the distant event t0. (Note that the moving observer does not regard these two events as simultaneous!)

OT and Ot0 are [timelike] legs of a [Minkowski]-right triangle on a spacetime diagram (as drawn below.... time runs upwards by convention).

The legs of the triangle are OT, Ot0 and Tt0.
Ot0 is the "hypotenuse" of this [Minkowski]-right triangle.
OT is the "adjacent side" and Tt0 is the "opposite side"... these legs are [Minkowski]-perpendicular.
Another way to describe this is that the vector Ot0 is being resolved into temporal and spatial components by our stationary observer.
[By the way, the "angle" $$\theta$$ is called the rapidity... and it is related to the relative velocity by $$v=c\tanh\theta$$. ]

So, think this way:

(adjacent side OT) = "COSINE(ANGLE)" (hypotenuse Ot0),
where "COSINE(ANGLE)" in the Minkowski geometry is $$\cosh\theta=\frac{1}{\sqrt{1-\tanh^2\theta}}=\frac{1}{\sqrt{1-(v/c)^2}}=\gamma$$

$$\begin{picture}(200,200)(0,0) \put(50,30){O} \put(50,50){\textcolor{red}{\line(3,4){52}}} \put(50,50){\textcolor{green}{\line(0,1){70}}} \put(50,120){\textcolor{green}{\line(1,0){50}}} \put(50,125){T} \put(100,125){$t_0$} \put(48,65){$\theta$} \end{picture}$$

A spacetime diagram is worth a thousand words.

7. Jun 7, 2005

### pmb_phy

That is quite true Rob. Someone asked a question here a while back which was best answered with a spacetime diagram. Here is the answer I gave to the question given at the top of the page

http://www.geocities.com/physics_world/sr/st_diagram.htm

I think the questioner was confusing spatial distances with spacetime intervals.

Pete

8. Jun 7, 2005

### robphy

It's a mystery to me why "spacetime diagrams" are rarely found in the relativity section of introductory textbooks.... but they're everywhere in the [Galilean] kinematics section! They really can clear up a lot of ambiguities and misunderstandings.

9. Jun 7, 2005

### pmb_phy

Good question. Let me see what I can learn on that. It would seem like a natural step since newer texts are introducing the spacetime interval and the spacetime diagram is awesome for describing that.

Pete

Last edited: Jun 7, 2005